 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-4/11i/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Pure periodicity means that

θ

reappears as the tail after one period. The continued-fraction tail formula therefore gives

θ = [a_{0}, \dots, a_{n}, θ] = \frac{p _{n} θ + p _{n - 1}}{q _{n} θ + q _{n - 1}} .

f (θ) = q_{n} θ^{2} + (q_{n - 1} - p_{n}) θ - p_{n - 1} = 0.

Now let

η = [\overline{a_{n}, a_{n - 1}, \dots, a_{0}}] .

By part (c), the matrix for one reversed period is

(p_{n} p_{n - 1} q_{n} q_{n - 1}),

so its positive fixed point

η

satisfies

p_{n - 1} η^{2} + (q_{n - 1} - p_{n}) η - q_{n} = 0.

Putting

η = - 1/ X

and multiplying by

- X^{2}

turns this equation into

f (X) = 0

. The product of the two roots of

f

- p_{n - 1} / q_{n} < 0

; because

θ > 0

, the other root

θ^{'}

is negative. Consequently

- 1/ θ^{'} > 0

- \frac{1}{θ ^{'}} = [\overline{a_{n}, a_{n - 1}, \dots, a_{0}}] .

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